It is well-known (I believe?) that the $p$-adics do not admit an ordering in the 'usual sense', the "usual sense" being a total order that is compatible with the field operations. I do not want to require that an ordering be a total order, however. In fact, I only require that it be a pre-order (though I believe it turns out to be a partial order).
Let $x\in \mathbb{Q}_p$ and define $x>0$ iff $x$ is a finite sum of powers of $p$ (the idea is that numbers which are non-zero absolute values should positive, and hence so should sums of them). I believe this should be a partial order on $\mathbb{Q}_p$ that is compatible with the field operations (i.e., $x\leq y$ implies $x+z\leq y+z$ and $x,y\geq 0$ implies $xy\geq 0$).
Does $\mathbb{Q}_p$ with this ordering have the least upper-bound property? (I apologize in advance if this question is overly easy; my understanding of the $p$-adics is not as strong as it probably should be.)
It does have the least upper-bound property.
Let $S\subseteq \mathbb{Q}_p$ be non-empty and bounded, so that there are $m,M\in \mathbb{Q}_p$ such that, if $x\in S$, then $m\leq x\leq M$. By replacing $S$ with $S-m$ we may without loss of generality assume that $m=0$. We then have that $0\leq x\leq M$.
Note that every non-zero element of $S$ is a finite sum of powers of $p$. In general, there are many ways of writing an element as a finite sum of powers of $p$, and so to aid our argument, we define a canonical way of writing each $x\in S$: when writing $x$ as a finite sum of powers of $p$, always group the terms into the largest powers of $p$ as possible. (For example, we write a sum of $p$ terms, each of which is $p^2$, as $p^3$.) Hereafter, we always write $x$ in canonical form as $x=a_mp^m+a_{m+1}p^{m+1}+a_{m+2}p^{m+2}+\cdots +a_np^n$ for $a_k\in \{ 0,\ldots ,p-1\}$.
If $M=0$, then because this is a partial order (this requires an easy check), it follows that $S=\{ 0\}$ and we are done, so we may as well suppose that $0\leq x<M$ for $x\in S$, in which case we can write $M=a_mp^m+\cdots +a_np^n$. Now if $x\in S$ is non-zero, we must have that $x=b_ip^i+\cdots +b_jp^j$, and also that $$ c_kp^k+\cdots +c_lp^l=M-x=(a_mp^m+\cdots +a_np^n)-(b_ip^i+\cdots +b_jp^j) $$ because $M-x>0$. Re-arranging, we must have $$ (b_ip^i+\cdots +b_jp^j)+(c_kp^k+\cdots +c_lp^l)=a_mp^m+a_np^n. $$ After re-writing the left-hand side in canonical form, for every power of $p$ between $p^i$ and $p^j$, either that power or $p$ times that power will remain (the power itself if we don't 'carry' or the power times $p$ if we do). In particular, every power of $p$ that appeared in $x$ must itself appear in $M$ or $p$ times that power must appear in $M$. Altogether, there are only finitely many possibilities for such a power.
In particular, this implies that $S$ is finite, and so, because the order restricted to $S$ is a total order (in fact, all integers are comparable with each other because one of the differences will be a finite sum of $p^0=1$s), we simply have that $\sup S=\max S$, and in particular, $S$ does have a least upper-bound.